Let's start with the two following equations and see what we can do here to get to a form we want:$PV=nRT$$D=\frac{m}{V}$We can also add another equation to the mix because we know that if we have a mass of a substance as well as a molar mass we can find the amount of substance:$n=\frac{m}{MM}$Let's rearrange this equation to solve for m, which we can shove into the density equation:$m=(n)(MM)$Now, if we substitute this into the density equation, we get:$D=\frac{(n)(MM)}{V}$Now, let's solve this for n, which we can shove into the ideal gas equation:$n=\frac{DV}{MM}$Now, finally let's plug this value into the ideal gas equation to get:$PV=nRT \rightarrow PV=\frac{DVRT}{MM}$The volumes on either side of the equation will cancel to give:$P=\frac{DRT}{MM}$We know pressure, density, the gas constant, and temperature, so we can solve for molar mass of the substance:$MM=\frac{DRT}{P} \rightarrow MM=\frac{(5.75\frac{g}{L})(0.08206\frac{L \times atm}{K \times mol})(298K)}{(1.00atm)}=141\frac{g}{mol}$And you'll notice that the units match up for a molar mass.